math calc help please

Question

i'm not sure how to approach these questions any help is appreciated!

1. g(x)= arcsin3x/x
2. g(x)=arccosx/x+1

Answer 1

g(x) = Arcsin(3x) / x

d/dx(Arcsinu) = [1/√(1-u²)] u'

So, g'(x) = [3x(1/√(1 - 9x²)) - Arcsin(3x)] / x²

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g(x) = Arccos(x) / (x+1)

d/dx(Arccosu) = [-1 / √(1 - u²)] u'

So, g'(x) = [-(x+1) / √(1 - x²) - Arccosx] / (x+1)²

^

Answer 2

Hi Ellie,

It is not quite clear what was the question they ask you.

However, I guess it might be about the difference in behavior of two functions near x=0

The function arcsin(3x) has a limit zero when x⇒0 in such a way that /x the ratio arsin(3x)/x has a final limit at x=0 (=3) [one cam use the L'opital Rule: lim [u(x)/v(x)] = lim [u'(x)/v'(x)] when x⇒0 to get it]

The function arcos(x) has final limit (Pi/2) at x=0 ; therefore arccos(x)/x would not have a limit (it goes to - ∞ when x⇒-0 and goes to + ∞ when x⇒+0)

The difference in heneral behavior between two functions is due to the fact that arsin(x) is an odf function of x while arccos(x) is an even function of x: arcsin(-x)=-arcsin(x) but arccos(-x) = + arccos(x)

Hope that'd help

Best,

Dr.Boroda